Problem: The side length of a square is increasing at a rate of $15$ millimeters per second. At a certain instant, the side length is $22$ millimeters. What is the rate of change of the area of the square at that instant (in square millimeters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $660$ (Choice B) B $484$ (Choice C) C $30$ (Choice D) D $225$
Setting up the math Let... $s(t)$ denote the square's side length at time $t$, and $A(t)$ denote the square's area at time $t$. We are given that $s'(t)=15$ and that $s(t_0)=22$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures $A(t)$ and $s(t)$ relate to each other through the formula for the area of a square: $A(t)=[s(t)]^2$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=2s(t)s'(t)$ Using the information to solve Let's plug ${s(t_0)}={22}$ and ${s'(t_0)}={15}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=2{s(t_0)}{s'(t_0)} \\\\ &=2({22})({15}) \\\\ &=660 \end{aligned}$ In conclusion, the rate of change of the area of the square at that instant is $660$ square millimeters per second. Since the rate of change is positive, we know that the area is increasing.